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a(n) = digital root of 2^Fibonacci(n).
0

%I #55 Dec 09 2024 11:01:08

%S 1,2,2,4,8,5,4,2,8,7,2,5,1,5,5,7,8,2,7,5,8,4,5,2,1,2,2,4,8,5,4,2,8,7,

%T 2,5,1,5,5,7,8,2,7,5,8,4,5,2,1,2,2,4,8,5,4,2,8,7,2,5,1,5,5,7,8,2,7,5,

%U 8,4,5,2,1,2,2,4,8,5,4,2,8,7,2,5,1,5,5,7,8,2,7,5,8,4,5,2

%N a(n) = digital root of 2^Fibonacci(n).

%C This sequence has period 24.

%H <a href="/index/Rec#order_24">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

%F a(n) = (a(n-1) * a(n-2) - 1) mod 9 + 1.

%F a(n) = A010888(A000301(n)).

%t Module[{a, n}, RecurrenceTable[{a[n] == Mod[a[n-1]*a[n-2] - 1, 9] + 1, a[0] == 1, a[1] == 2}, a, {n, 0, 99}]] (* or *)

%t PadRight[{}, 100, {1, 2, 2, 4, 8, 5, 4, 2, 8, 7, 2, 5, 1, 5, 5, 7, 8, 2, 7, 5, 8, 4, 5, 2}] (* _Paolo Xausa_, Nov 08 2024 *)

%Y Cf. A000301, A010888.

%K nonn,easy,base

%O 0,2

%A _Robert Bruce Gray_, Oct 20 2024